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根本不用矩阵和快速幂啊

Posted by qq648771054 at 2015-06-07 12:16:28 on Problem 3744
//根据递推公式观察数据发现,超过步数100以后的概率几乎等于一个和p有关的常数,
//所以可以直接认为走的步数大于100时的概率等于走到第99步的概率
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int cmp(int a,int b)
{
	return a<b;
}
int main()
{
	double p;
	int n;
	while(scanf("%d%lf",&n,&p)!=EOF)
	{
		double dp[100]={0};
		dp[1]=1;
		dp[0]=0;
		for(int i=2;i<=99;i++) dp[i]=dp[i-1]*p+dp[i-2]*(1-p);
		int i;
		int pos[20];
		for(i=0;i<n;i++)scanf("%d",pos+i);
		sort(pos,pos+n,cmp)	;
		int now=1;
		double r=1;
		for(i=0;i<n;i++)
		{
			
			if(pos[i]-now>=99) r*=dp[99]*(1-p);
			else r*=dp[pos[i]-now]*(1-p);
			now=pos[i]+1;
		}
		printf("%.7f\n",r);
	}
	
}

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